The reaction below is performed with 10.0 g of Al and 19.0 g of O2. How many grams of aluminum oxide will be made? 4 Al (s) + 3 O2 (g) → 2 Al2O3 (s)

18.89 g of Al₂O₃.

Explanation:

The balanced equation for the reaction is given below:

4Al + 3O₂ —> 2Al₂O₃

Next, we shall determine the masses of Al and O₂ that reacted and the mass of Al₂O₃ produced from the balanced equation. This can be obtained as follow:

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 4 × 27 = 108 g

Molar mass of O₂ = 16 × 2 = 32 g/mol

Mass of O₂ from the balanced = 3 × 32

= 96 g

Molar mass of Al₂O₃ = (27×2) + (16×3)

= 54 + 48

= 102 g/mol

Mass of Al₂O₃ from the balanced equation = 2 × 102 = 204 g

SUMMARY:

From the balanced equation above,

108 g of Al reacted with 96 g of O₂ to produce 204 g of Al₂O₃.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

108 g of Al reacted with 96 g of O₂.

Therefore, 10 g of Al will react with

= (10 × 96)/108 = 8.89 g of O₂.

From the calculation made above, we can see clearly that only 8.89 g out of 19 g of O₂ given, reacted completely with 10 g of Al.

Therefore, Al is the limiting reactant and O₂ is the excess reactant.

Finally, we shall determine the mass of aluminium oxide, Al₂O₃, produced from the reaction.

NOTE: in this case, the limiting reactant will be used because it will give the maximum mass of the aluminium oxide, Al₂O₃ as all of it is consumed in the reaction.

Al is the limiting reactant and the mass of aluminium oxide, Al₂O₃ produced can be obtained as follow:

From the balanced equation above,

108 g of Al reacted to produce 204 g of Al₂O₃.

Therefore, 10 g of Al will react to produce = (10 × 204)/108 = 18.89 g of Al₂O₃.

Thus, 18.89 g of Al₂O₃ were obtained from the reaction.

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